A sinusoidal wave travelling in the positive direction on a stretched string has amplitude `2.0 cm`, wavelength `1.0 m` and velocity `5.0 m//s`. At `x = 0` and ` t= 0` it is given that `y = 0` and `(dely)/(delt) lt 0`. Find the wave function `y (x, t)`.

Options

A. `Y(x,t)=(0.02 m)sin[(2pi m^(-1))x+(10pis^(-1))t]m`
B. `y(x,t)=(0.02 m)cos(10pis^(-1))t+(2pi m^(-1))xm`
C. `y(x,t0=(0.02 m)sin[(2pi m^(-1))x-(10pi s^(-1))t]m`
D. `y(x,t)=(0.02m)sin[(pim^(-1))x+(pis^(-1))t]m`

Answer

c

Explanation

We start with a general form for a rightward moving wave,
`y(x,t)=A sin(kx-omegat+phi)`
The amplitude given is `A=2.0 cm=0.02 m`.
The wavelength is given as,
`lambda=1.0 m`
wave number `=k=2pi//lambda=2pi m^(-1)`
Angular frequency,
`omega =vk =10pi rad //s`
`y(x,t)=(0.02) sin [2pi(x-5.0 t)+phi]`
we are told that for `x=0, t=0`,
`y=0` and `(dely)/(delt)lt0`
i.e., `0.02 sin phi =0` `(as y=0)`
and `-0.2 pi cos philt0`
From these condition, we may conclude that
`phi=2npi` where `n=0,2,4,6,......`
Therefore,
`y(x,t)=(0.02 m) sin [(2pi m^(-1))x-(10pi s^(-1))t] m`